Key Concepts and Formulas

• Sum of Series: If $S_n = \sum_{r=1}^n T_r$, then the $n$-th term of the series, $T_n$, can be found using the relation $T_n = S_n - S_{n-1}$ for $n > 1$. For $n=1$, $T_1 = S_1$.
• Partial Fraction Decomposition: A rational expression can be decomposed into simpler fractions, which is crucial for telescoping series. For an expression of the form $\frac{1}{(ax+b)(cx+d)(ex+f)}$, we can often use partial fractions to express it as a sum or difference of simpler terms.
• Telescoping Series: A series where intermediate terms cancel out, leaving only the first and last terms (or a finite number of terms). This simplifies the calculation of the sum.
• Limits of Sequences: Evaluating the behavior of a sequence as the number of terms approaches infinity.

Step-by-Step Solution

Step 1: Find the general term $T_n$ of the series. We are given the sum of the first $n$ terms, $S_n = \sum_{r=1}^n T_r = \frac{(2 n-1)(2 n+1)(2 n+3)(2 n+5)}{64}$. To find the general term $T_n$, we use the formula $T_n = S_n - S_{n-1}$ for $n > 1$.

First, let's write out $S_n$ and $S_{n-1}$: $S_n = \frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$ $S_{n-1} = \frac{(2(n-1)-1)(2(n-1)+1)(2(n-1)+3)(2(n-1)+5)}{64}$ $S_{n-1} = \frac{(2n-3)(2n-1)(2n+1)(2n+3)}{64}$

Now, we compute $T_n = S_n - S_{n-1}$: $T_n = \frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64} - \frac{(2n-3)(2n-1)(2n+1)(2n+3)}{64}$ We can factor out the common terms: $(2n-1)(2n+1)(2n+3)/64$. $T_n = \frac{(2n-1)(2n+1)(2n+3)}{64} [(2n+5) - (2n-3)]$ $T_n = \frac{(2n-1)(2n+1)(2n+3)}{64} [2n+5 - 2n+3]$ $T_n = \frac{(2n-1)(2n+1)(2n+3)}{64} [8]$ $T_n = \frac{8(2n-1)(2n+1)(2n+3)}{64}$ $T_n = \frac{(2n-1)(2n+1)(2n+3)}{8}$

Step 2: Find the reciprocal of the general term, $\frac{1}{T_n}$. We need to calculate $\lim_{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{T_r}$. So, we first find $\frac{1}{T_r}$. From Step 1, we have $T_r = \frac{(2r-1)(2r+1)(2r+3)}{8}$. Therefore, $\frac{1}{T_r} = \frac{8}{(2r-1)(2r+1)(2r+3)}$

Step 3: Express $\frac{1}{T_r}$ using partial fraction decomposition. The expression $\frac{8}{(2r-1)(2r+1)(2r+3)}$ can be decomposed. We observe that the denominators are in arithmetic progression with a common difference of 2. We can write: $\frac{8}{(2r-1)(2r+1)(2r+3)} = \frac{A}{2r-1} + \frac{B}{2r+1} + \frac{C}{2r+3}$

However, a more direct approach for terms like this, especially when looking for telescoping sums, is to consider differences of terms with two factors. Let's try to express $\frac{1}{(2r-1)(2r+1)(2r+3)}$ as a difference of two terms. Consider the expression $\frac{1}{(2r-1)(2r+1)} - \frac{1}{(2r+1)(2r+3)}$. $\frac{1}{(2r-1)(2r+1)} - \frac{1}{(2r+1)(2r+3)} = \frac{(2r+3) - (2r-1)}{(2r-1)(2r+1)(2r+3)}$ $= \frac{2r+3-2r+1}{(2r-1)(2r+1)(2r+3)} = \frac{4}{(2r-1)(2r+1)(2r+3)}$

So, we have $\frac{1}{(2r-1)(2r+1)(2r+3)} = \frac{1}{4} \left( \frac{1}{(2r-1)(2r+1)} - \frac{1}{(2r+1)(2r+3)} \right)$.

Now, substitute this back into the expression for $\frac{1}{T_r}$: $\frac{1}{T_r} = 8 \times \frac{1}{4} \left( \frac{1}{(2r-1)(2r+1)} - \frac{1}{(2r+1)(2r+3)} \right)$ $\frac{1}{T_r} = 2 \left( \frac{1}{(2r-1)(2r+1)} - \frac{1}{(2r+1)(2r+3)} \right)$

Step 4: Evaluate the sum $\sum_{r=1}^n \frac{1}{T_r}$. We need to compute $\sum_{r=1}^n \frac{1}{T_r} = \sum_{r=1}^n 2 \left( \frac{1}{(2r-1)(2r+1)} - \frac{1}{(2r+1)(2r+3)} \right)$. This is a telescoping series. Let's write out the first few terms: For $r=1$: $2 \left( \frac{1}{(2(1)-1)(2(1)+1)} - \frac{1}{(2(1)+1)(2(1)+3)} \right) = 2 \left( \frac{1}{1 \cdot 3} - \frac{1}{3 \cdot 5} \right)$ For $r=2$: $2 \left( \frac{1}{(2(2)-1)(2(2)+1)} - \frac{1}{(2(2)+1)(2(2)+3)} \right) = 2 \left( \frac{1}{3 \cdot 5} - \frac{1}{5 \cdot 7} \right)$ For $r=3$: $2 \left( \frac{1}{(2(3)-1)(2(3)+1)} - \frac{1}{(2(3)+1)(2(3)+3)} \right) = 2 \left( \frac{1}{5 \cdot 7} - \frac{1}{7 \cdot 9} \right)$ ... For $r=n$: $2 \left( \frac{1}{(2n-1)(2n+1)} - \frac{1}{(2n+1)(2n+3)} \right)$

Summing these terms: $\sum_{r=1}^n \frac{1}{T_r} = 2 \left[ \left(\frac{1}{1 \cdot 3} - \frac{1}{3 \cdot 5}\right) + \left(\frac{1}{3 \cdot 5} - \frac{1}{5 \cdot 7}\right) + \left(\frac{1}{5 \cdot 7} - \frac{1}{7 \cdot 9}\right) + \ldots + \left(\frac{1}{(2n-1)(2n+1)} - \frac{1}{(2n+1)(2n+3)}\right) \right]$

The intermediate terms cancel out. This is a telescoping sum. $\sum_{r=1}^n \frac{1}{T_r} = 2 \left[ \frac{1}{1 \cdot 3} - \frac{1}{(2n+1)(2n+3)} \right]$ $\sum_{r=1}^n \frac{1}{T_r} = 2 \left[ \frac{1}{3} - \frac{1}{(2n+1)(2n+3)} \right]$

Step 5: Evaluate the limit as $n \rightarrow \infty$. We need to find $\lim_{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{T_r}$. $\lim_{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{T_r} = \lim_{n \rightarrow \infty} 2 \left[ \frac{1}{3} - \frac{1}{(2n+1)(2n+3)} \right]$

As $n \rightarrow \infty$, the term $\frac{1}{(2n+1)(2n+3)}$ approaches 0 because the denominator grows infinitely large. $\lim_{n \rightarrow \infty} \frac{1}{(2n+1)(2n+3)} = 0$

Therefore, the limit is: $\lim_{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{T_r} = 2 \left[ \frac{1}{3} - 0 \right]$ $\lim_{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{T_r} = 2 \times \frac{1}{3} = \frac{2}{3}$

Common Mistakes & Tips

• Incorrectly calculating $T_n$: Ensure that the subtraction $S_n - S_{n-1}$ is performed accurately, paying close attention to signs and common factors.
• Errors in Partial Fraction Decomposition: For telescoping series, it's often easier to look for a difference of two terms that result in the desired numerator when combined, rather than a full three-term partial fraction decomposition.
• Forgetting the Constant Factor: The factor of 8 in $T_n$ and the subsequent factor of 2 in $\frac{1}{T_r}$ are crucial. Make sure they are carried through the calculation.
• Incorrectly identifying the telescoping pattern: Visually write out a few terms of the sum to confirm that intermediate terms are indeed canceling out.

Summary

The problem requires finding the limit of the sum of the reciprocals of the terms of a series. First, we determined the general term $T_n$ of the series by subtracting the sum of the first $n-1$ terms ($S_{n-1}$) from the sum of the first $n$ terms ($S_n$). This yielded $T_n = \frac{(2n-1)(2n+1)(2n+3)}{8}$. Next, we found the reciprocal of the general term, $\frac{1}{T_r} = \frac{8}{(2r-1)(2r+1)(2r+3)}$. Using partial fraction decomposition in a specific form suitable for telescoping sums, we expressed $\frac{1}{T_r}$ as $2 \left( \frac{1}{(2r-1)(2r+1)} - \frac{1}{(2r+1)(2r+3)} \right)$. Summing this expression from $r=1$ to $n$ resulted in a telescoping series whose sum is $2 \left[ \frac{1}{3} - \frac{1}{(2n+1)(2n+3)} \right]$. Finally, we evaluated the limit of this sum as $n \rightarrow \infty$, which simplified to $\frac{2}{3}$.

The final answer is \boxed{\frac{2}{3}}. This corresponds to option (A).